Find the current “I” by using the superposition theorem.

Hint : In this question, we need to determine the value of ‘I’ by following the superposition theorem only. For this, we will take one source at a time and evaluate the value of the current passing through the resistance of 2 ohms.

Complete Step By Step Answer:
According to the Superposition theorem, in a circuit containing more than one independent source, the current through or voltage across an element is the sum of the effect caused due to sources acting one at a time.
Let us take the effect of 5 volts first so the 10 volts side will be short-circuited.

Here, 4 ohms and 6 ohms are connected in parallel. And upper 2 ohms resistance is connected in parallel with the series combination of lower 2 ohms and (4 ll 6).
The circuit diagram will be reduced to

A voltage of 5 volts is applied to the upper 2 ohms resistance. So, following ohms’ law, we get
$
  {I_2} = \dfrac{5}{2} \\
   = 2.5A(clock - wise) - - - - (i) \\
 $
Now, considering 10 ohms resistance to be acting alone.

Here, lower 2 ohms and 6 ohms are connected in parallel. And upper 2 ohms resistance is connected in parallel with the series combination of lower 4 ohms and (2 ll 6).
The circuit diagram will be reduced to

A voltage of 10 volts is applied to the upper 2 ohms resistance. So, following ohms’ law, we get
$
  {I_2} = \dfrac{{10}}{2} \\
   = 5A(anticlock - wise) - - - - (ii) \\
 $
By equation (i) and (ii), we can say that the equivalent current flowing through the resistance of 2 ohms is
$
  I = 2.5 - 5 \\
   = - 2.5(clockwise) \\
 $
Or,
$
  I = 5 - 2.5 \\
   = 2.5(anticlockwise) \\
 $
Hence, the value of “I” is 2.5 Amperes in an anti-clockwise direction.

Note :
When a voltage source is zero then, the circuit is short-circuited, and when the current source is made zero then, the circuit is open-circuited. It should be noted here that the direction of the current plays a vital role in determining the net current through the resistance.